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Chapter 9 : Linear Momentum
Study Guide is posted online in the HW section Exam 3 is on Nov. 19
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9-7 Elastic Collisions: 2D
Momentum is conserved vAx vAy v’Ax v’Ay V’Bx v’by vBy vBx Air bags Egg drop Rocket engines X-direction: Y-direction: Kinetic Energy is also conserved
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9-7 Inelastic Collisions: 2D
Momentum is conserved vAx vAy v’x v’y vBx vBy X-direction Y-direction Air bags Egg drop Rocket engines Objects stick together
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9-7 Collisions in Two or Three Dimensions
Conservation of energy and momentum can also be used to analyze collisions in two or three dimensions, but unless the situation is very simple, the math quickly becomes unwieldy. Here, a moving object collides with an object initially at rest. Knowing the masses and initial velocities is not enough; we need to know the angles as well in order to find the final velocities. Figure Caption: Object A, the projectile, collides with object B, the target. After the collision, they move off with momenta pA’ and pB’ at angles θA’ and θB’. The objects are shown here as particles, as we would visualize them in atomic or nuclear physics. But they could also be macroscopic pool balls.
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9-7 Collisions in Two or Three Dimensions
Example 9-12: Billiard ball collision in 2-D. Billiard ball A moving with speed vA = 3.0 m/s in the +x direction strikes an equal-mass ball B initially at rest. The two balls are observed to move off at 45° to the x axis, ball A above the x axis and ball B below. That is, θA’ = 45° and θB’ = 45 °. What are the speeds of the two balls after the collision? Figure 9-19. Solution: Apply conservation of momentum; the masses are the same, as are the outgoing angles. The final speeds are equal; both are 2.1 m/s.
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9-8 Center of Mass (CM) In (a), the diver’s motion is pure translation; in (b) it is translation plus rotation. There is one point that moves in the same path a particle would take if subjected to the same force as the diver. This point is called the center of mass (CM). Figure Caption: The motion of the diver is pure translation in (a), but is translation plus rotation in (b). The black dot represents the diver’s CM at each moment.
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9-8 Center of Mass (CM) The general motion of an object can be considered as the sum of the translational motion of the CM, plus rotational, vibrational, or other forms of motion about the CM. Figure Caption: Translation plus rotation: a wrench moving over a horizontal surface. The CM, marked with a red cross, moves in a straight line.
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9-8 Center of Mass (CM) For two particles, the center of mass lies closer to the one with the most mass: where M is the total mass. Figure Caption: The center of mass of a two-particle system lies on the line joining the two masses. Here mA > mB, so the CM is closer to mA than to mB .
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9-8 Center of Mass (CM) Example 9-14: CM of three guys on a raft.
Three people of roughly equal masses m on a lightweight (air-filled) banana boat sit along the x axis at positions xA = 1.0 m, xB = 5.0 m, and xC = 6.0 m, measured from the left-hand end. Find the position of the CM. Ignore the boat’s mass. Figure 9-24. Solution: For more than two particles, finding the center of mass is an extension of the two-particle process; just add the product of each mass and its distance from some reference point and divide by the total mass. This gives the distance of the CM from the reference point. Here it is 4.0 m.
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Problem 62 Problem 62:The cm of an empty 1250-kg car is 2.50 m behind the front of the car. How far from the front of the car will the cm be when two people sit in the front seat 2.80 m from the front of the car, and three people sit in the back seat 3.90 m from the front? Assume that each person has a mass of 70.0 kg.
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9-8 Center of Mass (CM) Exercise 9-15: Three particles in 2-D.
Three particles, each of mass 2.50 kg, are located at the corners of a right triangle whose sides are 2.00 m and 1.50 m long, as shown. Locate the center of mass. Figure 9-25. Solution: This is the same process as before, except that we have to do it separately for the x and y components of the position of the center of mass. We find x = 1.33 m and y = 0.50 m.
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9-8 Center of Mass (CM) For an extended object, we imagine making it up of tiny particles, each of tiny mass, and adding up the product of each particle’s mass with its position and dividing by the total mass. In the limit that the particles become infinitely small, this gives: Figure Caption: An extended object, here shown in only two dimensions, can be considered to be made up of many tiny particles (n), each having a mass Δmi. One such particle is shown located at a point ri = xii + yij + zik. We take the limit of n →∞ so Δmi becomes the infinitesimal dm.
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9-8 Center of Mass (CM) The center of gravity is the point at which the gravitational force can be considered to act. It is the same as the center of mass as long as the gravitational force does not vary among different parts of the object. Figure Caption: Determining the CM of a flat uniform body.
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9-9 Center of Mass and Translational Motion
The total momentum of a system of particles is equal to the product of the total mass and the velocity of the center of mass. The sum of all the forces acting on a system is equal to the total mass of the system multiplied by the acceleration of the center of mass: Therefore, the center of mass of a system of particles (or objects) with total mass M moves like a single particle of mass M acted upon by the same net external force.
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